n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an angle θ, of 60° with the floor. A man weighing 864 N climbs slowly up the ladder When he has climbed to a point that is 7.8 m from the base of the ladder, the ladder starts to slip. What is the coefficient of static friction between the floor and the ladder?

Question

anonymous · Answer

I can't get the coefficient of static friction

anonymous · Answer

that's what the problem is asking for...

anonymous · Answer

did you do sum of forces/torques ?

anonymous · Answer

I got confused in summing up the torques

anonymous · Answer

and also in summing the forces.

anonymous · Answer

I chose rotation about the point of contact with the floor.
that gives you an equation with the normal force between the wall and the ladder...
you have two equations from sum of forces...  so you can solve for / eliminate both normal forces and the coefficient...

anonymous · Answer

what did you get for sum of 'x' forces?

anonymous · Answer

horizontal  forces...

anonymous · Answer

hello?

anonymous · Answer

the sum of x forces=friction + normal force in x =0

anonymous · Answer

yep ok, I guess you have made an attempt... so I'll draw you a sketch that should clarify what your 3 eqn.s should be...

anonymous · Answer

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anonymous · Answer

vectors not to scale...

anonymous · Answer

N2 - us*N1 = 0
N1 - 864 -400 =0

anonymous · Answer

use the point of contact with the floor as your axis of rotation...

anonymous · Answer

torque from the 400N force will be...?

anonymous · Answer

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